A gas mixture contains 1.204x10^23 molecules of H2 and 7.1 g of Cl2 (Cl = 35.5). Using N_A = 6.02x10^23 mol^-1 and 22.4 L mol^-1 at STP, find the volume of the mixture at STP in litres (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3mc010

A gas mixture contains $1.204 \times 1 0^{23}$ molecules of $H X_{2}$ and $7.1 g$ of $Cl X_{2}$ (Cl = 35.5). Using $N_{A} = 6.02 \times 1 0^{23} mo l^{- 1}$ and $22.4 L mo l^{- 1}$ at STP, find the volume of the mixture at STP in litres (two decimals).

Numerical answer type, answer in L, tolerance 0.005.

n (H X_{2}) = 1.204 \times 1 0^{23} /6.02 \times 1 0^{23} = 0.20

mol;

n (Cl X_{2}) = 7.1/71 = 0.10

mol. Total

0.30

mol, so

V = 0.30 \times 22.4 = 6.72 L